limit of product of $(a_1a_2.\dots a_n)^{\frac{1}{n}}$
You should recognize the logarithm as
$$\frac{1}{n} \sum_{k = 1}^n \ln\left(1 + \frac k n\right)$$
which is a Riemann sum for $\int_1^2 \ln x \, dx$.
As far as general rules, things are frequently a bit ad hoc. If you can find a Riemann sum, it's helpful. That's frequently going to happen precisely because of the factor $1/n$ that you can pull down with a log.
As can be seen if $$a_{n} = \frac{(n + 1)(n + 2)\cdots (2n)}{n^{n}}$$ then the sequence in question is $a_{n}^{1/n}$. Now we can see that $$\frac{a_{n + 1}}{a_{n}} = \frac{(n + 2)(n + 3)\cdot (2n)(2n + 1)(2n + 2)}{(n + 1)^{n + 1}}\cdot\frac{n^{n}}{(n + 1)(n + 2) \cdots (2n)}$$ so that $$\frac{a_{n + 1}}{a_{n}} = \frac{2(2n + 1)}{n + 1}\cdot\left(\frac{n}{n + 1}\right)^{n} \to \frac{4}{e}$$ as $n \to \infty$ Hence the sequence $a_{n}^{1/n}$ also tends to $4/e$.
In general if $b_{n} = (a_{1}a_{2}\cdots a_{n})^{1/n}$ and $a_{n} \to L$ then $b_{n} \to L$.
Taking the $\log$ of the expression, we arrive to the sum
$$ \frac{1}{n} \sum_{k=1}^n \log \left( 1 + \frac{k}{n} \right). $$
This sum is a Riemann sum for $\int_1^2 \log(x) \, dx = \left[ x\log x - x\right]_{x=1}^{x=2} = 2\log(2) - 1$. Hence, the limit is
$$ e^{2\log(2) - 1} = \frac{4}{e}.$$